Question

Use the quadratic formula to solve x^2 + 9x + 10 = 0. Estimate irrational solutions to the nearest tenth.

Options.
{–3.6, 16.4}
{–7.7, –1.3}
{–15.4, –2.6}
{–25, 16}

Answer 1

try pluging in the points

Answer 2

you should get the awnser

Answer 3

@rational

Answer 4

@jim_thompson5910

Answer 5

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

\[\Large x=\frac{-9 \pm \sqrt{9^2-4*1*10}}{2*1}\]

\[\Large x=\frac{-9 \pm \sqrt{41}}{2}\]

I'll let you finish up

Answer 6

option c?

Answer 7

Actually, this equation is easy enough to solve in your head to find the answer choice!

You can also check your answer (it is incorrect) by plugging into the formula.

if \(x=-2.6\) as your answer choice suggests, then
\[x^2+9x+10=0\]or\[(-2.6)(-2.6)+9(-2.6)+10=0\]\[6.76-23.4+10=0\]Uh, no, that's not true, so that is not a correct answer.

Answer 8

Jim worked the entire problem for you, except evaluating it to a decimal. Can you tell me how you came up with option c?

Answer 9

becuase

Answer 10

While that is true, that's not the entire formula he gave you — you have to divide that by 2.

Answer 11

Option C is what I like to call a sucker answer — looks like a value you'll get at an intermediate point in the computation, or a value you'll get if you make a frequent mistake. You see something on your paper or calculator that matches one of the answers and say "that must be it!" without bothering to check if it actually fits the problem.

Answer 12

oh, i see so it must be option B.

Answer 13

Thank you for your help

Answer 14

:)

Answer 15

By the way, you can work this one out pretty easily in your head.

\[x^2+9x+10=0\]For the moment, let's pretend that "\(+10\)" isn't there.

\[x^2+9x\cancel{+10}=0\]\[x^2+9x=0\]\[x^2=-9x\]\[\frac{x^2}{x} = \frac{-9x}{x}\]\[x=-9\]So we can guess that a value of \(x\) that satisfies the equation is somewhere in the neighborhood of -9. We can proceed in a number of ways from here to tighten up our estimate. First is to plug that value in for \(x\) and see how close it is:

\[(-9)^2 + 9(-9) + 10 = 0\]\[81-81+10=0\]Uh, well, not exactly :-) The \(x^2\) term is positive for all values of \(x\), so we need to make the first term + 10 equal the second term (except the second term will be negative). That means we need the value of \(x^2\) to be about \(10\) less than it is now (81). \(81-10 = 71\) What is a number which multiplied by itself gives about \(71\)? \(9*9 = 81\), how about \(8*8 = 64\)? Let's try \(-8\) in our equation:

\[(-8)*(-8) + 9(-8) + 10 = 0\]
\[64-72+10 = 0\]\[-8+10=0\]We are getting closer! How about trying \(-7\):
\[(-7)*(-7)+9(-7) + 10 = 0\]\[49-63+10 = 0\]\[-14+10=0\]This is not as close as the previous try, so we must have gone too far. But, we now know it is somewhere between \(-7\) and \(-8\), and if we look at our answer choices, there's only one that fits. We plug both of those answer choices into the formula to make sure that they do work and are not a trap.